Thin lens: the formula and the derivation of the formula. Solving problems with the thin lens formula
Now we are talking about geometric optics. In this section, much time is devoted to an object such as a lens. After all, it can be different. In this case, the formula of a thin lens is one for all cases. Only you need to know how to apply it correctly.
Types of lenses
It is always a body transparent to light rays, which has a special shape. The appearance of the object dictates two spherical surfaces. One of them can be replaced by a flat one.
And the lens may be thicker than the middle orthe edges. In the first case, it will be called convex, in the second case it will be concave. And, depending on how the concave, convex and flat surfaces are combined, the lenses can also be different. Namely: biconvex and biconcave, planeconvex and flatconcave, convexconcave and concaveconvex.
Under normal conditions, these objects are used inair. They are made from a substance whose optical density is greater than that of air. Therefore, the convex lens will be collecting, and the concave lens will dissipate.
General characteristics
Before talking aboutthin lens formula, you need to decide on the basic concepts. They need to know. Because they will constantly handle a variety of tasks.
The main optical axis is a straight line. It is drawn through the centers of both spherical surfaces and determines the location of the center of the lens. There are still additional optical axes. They are drawn through a point that is the center of the lens, but do not contain centers of spherical surfaces.
In the formula of a thin lens there is a quantity that determines its focal length. So, the focus is the point on the main optical axis. It intersects rays that run parallel to this axis.
And the foci of every thin lens is always two. They are located on both sides of its surfaces. Both focuses have a valid focus. In the scattering  imaginary.
The distance from the lens to the point of focus is the focal length (letterAnd its value can be positive (in the case of a collecting one) or negative (for scattering).
With the focal length, another characteristic is associated: the optical force. It is customary to designateIts value is always the inverse of the focus, that is,The optical force in diopters is measured (abbreviated, dpt).
What other notation is in the formula of a thin lens
In addition to the already indicated focal length, it will be necessary to know several distances and dimensions. For all types of lenses they are the same and are presented in the table.
Notation  Name 
d  distance to the object 
h  height of the studied subject 
f  distance to image 
H  height of the resulting image 
All specified distances and heights are usually measured in meters.
In physics with the formula of a thin lens, the concept of magnification is also connected. It is defined as the ratio of the image dimensions to the height of the object, that is, H / hIt can be denoted by G.
What you need to build an image in a thin lens
It is necessary to know this in order to obtain the formulathin lens that collects or diffuses. The drawing begins with the fact that both lenses have their own schematic image. Both of them look like a piece. Only at the arrows that gather at its ends are directed outward, and at the scattering points  inside this segment.
Now to this segment it is necessary to carry outperpendicular to its middle. This will show the main optical axis. On it on both sides of the lens at the same distance, it is necessary to note foci.
The object, the image of which is to be constructed, is drawn in the form of an arrow. It shows where the top of the object is. In general, the object is placed parallel to the lens.
How to build an image in a thin lens
In order to construct an image of the object,it is enough to find the points of the ends of the image, and then to connect them. Each of these two points can be obtained from the intersection of two rays. The simplest in construction are two of them.

Going from the specified point in parallel to the mainoptical axis. After contact with the lens, it goes through the main focus. If we are talking about a collecting lens, then this focus is behind the lens and the beam goes through it. When the scattering is considered, the beam must be drawn so that its continuation passes through the focus in front of the lens.

Going directly through the optical center of the lens. He does not change his direction for her.
There are situations when the subject is deliveredperpendicular to the main optical axis and ends on it. Then it is sufficient to construct an image of the point that corresponds to the edge of the arrow not lying on the axis. And then draw from it a perpendicular to the axis. This will be the image of the object.
The intersection of the constructed points gives an image. In a thin collecting lens, a real image is obtained. That is, it is obtained directly at the intersection of the rays. An exception is the situation where an object is placed between the lens and the focus (as in a magnifying glass), then the image is imaginary. For scattering it always turns out to be imaginary. After all, it is obtained at the intersection of not the rays themselves, but their extensions.
Actual image is usually drawn with a solid line. But the imaginary one is dotted. This is due to the fact that the first is actually there, and the second only is seen.
The derivation of the formula of a thin lens
This is conveniently done on the basis of a drawing illustrating the construction of a valid image in a collecting lens. The designation of the segments is indicated on the drawing.
The optics section is not for nothing called geometric. You will need knowledge from this section of mathematics. First, we need to consider the triangles AOB and A. They are similar, because they have two equal angles (straight and vertical). From their similarity it follows that the moduli of the segments Aand AB are referred to as modules of segments OBand OB.
Similar (on the basis of the same principle at two angles) are two more triangles:. In them the ratios of already modular segments are equal: Awith SB andProceeding from the construction, the segments AB and CD will be equal. Therefore, the lefthand sides of these relations are the same. Therefore, the right are equal. That is, OB/ OM is equal to
In this equality, the segments denoted by the points can be replaced by the corresponding physical concepts. So the IAIs the distance from the lens to the image. OB is the distance from the object to the lens.focal length. A segmentis equal to the distance difference between the image and the focus. Therefore, it can be rewritten in a different way:
Ff = df  dF.
To derive the formula of a thin lens, the last equation must be divided intoThen it turns out:
1 / d + 1 / f = 1 / F.
This is the formula of a thin collecting lens. The focal length of the scattering lens is negative. This leads to a change in the equality. True, it is insignificant. Just in the formula of a thin diffusing lens there is a minus before the ratio 1 /
1 / d + 1 / f =  1 / F.
The problem of finding the lens magnification
Condition.The focal length of the collecting lens is 0.26 m. It is required to calculate its magnification if the object is at a distance of 30 cm.
Decision. It starts with the introduction of notation and the translation of units in C. So, we know= 30 cm = 0.3 m and= 0.26 m. Now we need to choose the formulas, the main one is the one indicated for magnification, the second for the thin collecting lens.
They need to somehow unite. To do this, you will have to consider the drawing of the image construction in the collecting lens. From these triangles we see that Γ = H / h= f / d. That is, in order to find the increase, it is necessary to calculate the ratio of the distance to the image to the distance to the object.
The second is known. But the distance to the image is supposed to be derived from the formula indicated earlier. It turns out that
Now these two formulas must be combined.
At this point, the solution of the problem for the thin lens formula reduces to elementary calculations. It remains to substitute the known quantities:
T = 0.26 / (0.30.26) = 0.26 / 0.04 = 6.5.
The answer: the lens gives an increase of 6.5 times.
The task in which to find the focus
Condition.The lamp is located one meter from the collecting lens. The image of its spiral is obtained on the screen, separated from the lens by 25 cm. Calculate the focal length of this lens.
Decision.Data are supposed to write down such quantities:= 1 m and= 25 cm = 0.25 m. This information is sufficient to calculate the focal length from the formula of a thin lens.
= 1/1 + 1 / 0,25 = 1 + 4 = 5. But in the task you need to know the focus, not the optical power. Therefore, it remains only to divide 1 by 5, and we get the focal length:
Answer: The focal length of the collecting lens is 0.2 m.
The problem of finding the distance to the image
Condition. The candle was placed at a distance of 15 cm from the collecting lens. Its optical strength is 10 dpt. The screen behind the lens is set so that it produces a clear image of the candle. What is the distance equal to?
Decision.In a short entry it is supposed to write down such data:= 15 cm = 0.15 m,= 10 dpt. The formula derived above should be written with a slight change. Namely, put the righthand side of the equalityinstead of 1 /
After several transformations, a formula is obtained for the distance from the lens to the image:
Now it is necessary to substitute all the numbers and count them. This is the value for0.3 m.
Answer: The distance from the lens to the screen is 0.3 m.
The problem of the distance between an object and its image
Condition.The object and its image are separated by 11 cm. The collecting lens gives an increase of 3 times. Find its focal length.
Decision.The distance between the object and its image is conveniently denoted by the letter= 72 cm = 0.72 m. The increase in T = 3.
Here two situations are possible. The first  the object is behind the focus, that is, the image is real. In the second, the subject is between the focus and the lens. Then the image on the same side as the object, and imaginary.
Let's consider the first situation. The object and the image are on different sides of the collecting lens. Here you can write this formula:The second equation is to write: Γ =It is necessary to solve the system of these equations with two unknowns. To do this, replaceon 0,72 m, and Г on 3.
It follows from the second equation thatThen the first is transformed so: 0,72 = 4It is easy to calculate from it18 (m). Now it is easy to determine= 0.54 (m).
It remains to use the formula of a thin lens to calculate the focal length.= (0.18 * 0.54) / (0.18 + 0.54) = 0.135 (m). This is the answer for the first case.
In the second situation, the image is imaginary, and the formula forwill be different:The second equation for the system will be the same. Similarly, arguing, we get that36 (m), and= 1.08 (m). Such a calculation of the focal length will give the following result: 0.54 (m).
Answer: the focal length of the lens is 0.135 m or 0.54 m.
Instead of concluding
The course of rays in a thin lens is an importantpractical application of geometric optics. After all, they are used in many devices from a simple magnifying glass to accurate microscopes and telescopes. Therefore, it is necessary to know about them.
The deduced formula of a thin lens allows one to solveset of tasks. And it allows you to draw conclusions about what image gives different types of lenses. It is enough to know its focal length and distance to the object.